/*
 * P2470.cpp
 *
 *  Created on: Nov 28, 2012
 *      Author: yixiao

Description

Some programming contest problems are really tricky: not only do they require a different output format from what you might have expected, but also the sample output does not show the difference. For an example, let us look at permutations.
A permutation of the integers 1 to n is an ordering of these integers. So the natural way to represent a permutation is to list the integers in this order. With n = 5, a permutation might look like 2, 3, 4, 5, 1.
However, there is another possibility of representing a permutation: You create a list of numbers where the i-th number is the position of the integer i in the permutation. Let us call this second possibility an inverse permutation. The inverse permutation for the sequence above is 5, 1, 2, 3, 4.
An ambiguous permutation is a permutation which cannot be distinguished from its inverse permutation. The permutation 1, 4, 3, 2 for example is ambiguous, because its inverse permutation is the same. To get rid of such annoying sample test cases, you have to write a program which detects if a given permutation is ambiguous or not.
Input

The input contains several test cases.
The first line of each test case contains an integer n (1 <= n <= 100000). Then a permutation of the integers 1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between 1 and n appears exactly once in the permutation.
The last test case is followed by a zero.
Output

For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.
Sample Input

4
1 4 3 2
5
2 3 4 5 1
1
1
0

Sample Output

ambiguous
not ambiguous
ambiguous
Hint

Huge input,scanf is recommended.
 */

#include<iostream>
using namespace std;

int main(){
	int n,i;
	int m[100005];
	bool amb;
	while(true){
		scanf("%d",&n);
		if(n==0)
			break;
		for(i=0;i<n;i++){
			scanf("%d",m+i);
		}
		amb=true;
		for(i=0;i<n;i++){
			if(m[m[i]-1]!=i+1){
				amb=false;
				break;
			}
		}
		if(amb){
			printf("ambiguous\n");
		}else{
			printf("not ambiguous\n");
		}
	}
	return 0;
}


